Answer:
5123.79 kg
Explaination:
I did 5,123,788 divided by 1000 and got 5123.788
Then I rounded to the nearest 10th place.
Hope this helps, have a great day!
Answer:
The magnitude of the average induced emf in the wire during this time is 9.533 V.
Explanation:
Given that,
Radius r= 0.63 m
Magnetic field B= 0.219 T
Time t= 0.0572 s
We need to calculate the average induce emf in the wire during this time
Using formula of induce emf



.....(I)
In reshaping of wire, circumstance must remain same.
We calculate the length when wire is in two loops



The length when wire is in one loop




We need to calculate the initial area

Put the value into the formula


The final area is



Put the value of initial area and final area in the equation (I)


Negative sign shows the direction of induced emf.
Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.
The students can feel the vibrations from the bleachers, but not from the drums because of the resonance.
<h3>What is the resonance?</h3>
The resonance condition occurs when the frequency of a wave matches with the resonance frequency.
The transfer of energy becomes maximum when the natural frequency matches with the frequency generated by the source.
The bleacher and student together having the frequency matching with the resonance frequency of the drum.
The bleacher and student with the bag frequency doesn't match with the resonance value of drums. Secondly, as distance increases sound intensity also decreases.
Thus, the student can feel the vibrations from the bleachers, but not from the drums because of the resonance.
Learn more about resonance.
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